square roots and negative infinity

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October 25th 2009, 03:30 PM
Evan.Kimia

square roots and negative infinity

problem:
Find the limit. (If you need to use -http://www.webassign.net/images/infinity.gif or http://www.webassign.net/images/infinity.gif, enter -INFINITY or INFINITY.)
http://www.webassign.net/cgi-bin/sym...%204%20x%29%29

I would explain what ive done in more detail, but i know im missing something due to x approaching neg. infinity, but i did multiply the above funct. by its conjugate to get (x-1)/(x-square root of x^2+4x). How would i go about this? Thank you. -Evan
October 26th 2009, 04:44 AM
mr fantastic

Quote:

Originally Posted by Evan.Kimia

problem:
Find the limit. (If you need to use -http://www.webassign.net/images/infinity.gif or http://www.webassign.net/images/infinity.gif, enter -INFINITY or INFINITY.)
http://www.webassign.net/cgi-bin/sym...%204%20x%29%29

I would explain what ive done in more detail, but i know im missing something due to x approaching neg. infinity, but i did multiply the above funct. by its conjugate to get (x-1)/(x-square root of x^2+4x). How would i go about this? Thank you. -Evan

Let $x = -t$ and calculate $\lim_{t \rightarrow + \infty} \left( \sqrt{ t^2 - 4t} - t\right)$ .
October 26th 2009, 03:50 PM
Evan.Kimia

Ok here is what i have...

sqroot(t^2-4t)-t * sqroot(t^2-4t)+t
------------------
1 * sqroot(t^2-4t)+t

t^2-4t-t^2
-------------
sqroot(t^2-4t)+t

consolidate like terms then divide by t....
-4
---------------
* sqroot(t^2-4t)+1

how can i properly solve for the denominator with the sqroot involved? I have something on paper that gets me sqroot(t-4)+1 which is most likely wrong. Thanks.
October 26th 2009, 05:47 PM
mr fantastic

Quote:

Originally Posted by Evan.Kimia

Ok here is what i have...

sqroot(t^2-4t)-t * sqroot(t^2-4t)+t
------------------
1 * sqroot(t^2-4t)+t

t^2-4t-t^2
-------------
sqroot(t^2-4t)+t

consolidate like terms then divide by t....
-4
---------------
* sqroot(t^2-4t)+1

how can i properly solve for the denominator with the sqroot involved? I have something on paper that gets me sqroot(t-4)+1 which is most likely wrong. Thanks.

When you divided the denominator by t, you failed to divide the square root term by t. The correct expression is $\frac{-4}{\sqrt{1 - \frac{4}{t}} + 1}$ . Now take the required limit.
October 26th 2009, 06:02 PM
Evan.Kimia

I see. How would i properly divide t by the denominator? I know this is easy, but my algebra is off. I know dividing by t in the sqroot wont really work unless t itself is within a square root, so t= sqroot(t^2), am i right? If thats so, then i can see sqroot(t^2-4t)
turning into (1-4/t), but wouldnt the 1 turn into 1/sqroot(x^2) which would equal 0? I know im wrong, just thinking out loud.

Also, you explained this better than my math teacher today, thanks for everything.
October 26th 2009, 06:07 PM
mr fantastic

Quote:

Originally Posted by Evan.Kimia

I see. How would i properly divide t by the denominator? I know this is easy, but my algebra is off. I know dividing by t in the sqroot wont really work unless t itself is within a square root, so t= sqroot(t^2), am i right? If thats so, then i can see sqroot(t^2-4t)
turning into (1-4/t), but wouldnt the 1 turn into 1/sqroot(x^2) which would equal 0? I know im wrong, just thinking out loud.

Also, you explained this better than my math teacher today, thanks for everything.

You divide each term in the denominator by t.

Dividing the square root term by t works in the way you have described. And when you divide the other term by t (and that other term is t) then you get t/t = 1. So you end up with a denominator as I posted.